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Question Number 21772 by j.masanja06@gmail.com last updated on 03/Oct/17

$$integrate$$$$\int 2^{4x}dx$$

Answered by mrW1 last updated on 03/Oct/17

$$\int 2^{4\mathrm{x}}\mathrm{dx}$$$$=\frac{1}{4}\int 2^{4\mathrm{x}}\mathrm{d}\left(4\mathrm{x}\right)$$$$=\frac{1}{4}\int 2^{\mathrm{t}}\mathrm{dt}\mathrm{with}\mathrm{t}=4\mathrm{x}$$$$=\frac{1}{4}\int {\mathrm{e}}^{\mathrm{tln}2}\mathrm{dt}$$$$=\frac{1}{4\ln 2}\int {\mathrm{e}}^{\mathrm{tln}2}\mathrm{d}\left(\mathrm{tln}2\right)$$$$=\frac{1}{4\ln 2}\int {\mathrm{e}}^{\mathrm{u}}\mathrm{du}\mathrm{with}\mathrm{u}=\mathrm{tln}2=\ln 2^{\mathrm{t}}=\ln 2^{4\mathrm{x}}$$$$=\frac{1}{4\ln 2}\times {\mathrm{e}}^{\mathrm{u}}+\mathrm{C}$$$$=\frac{1}{4\ln 2}\times {\mathrm{e}}^{\ln 2^{4\mathrm{x}}}+\mathrm{C}$$$$=\frac{2^{4\mathrm{x}}}{4\ln 2}+\mathrm{C}$$

Answered by sma3l2996 last updated on 03/Oct/17

$$=\int e^{4xln2}dx=\frac{2^{4x}}{4ln2}+C$$

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